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gain bandwidth product of unity gain opamp

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issac newton

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i have attached a bode plot and the opamp dc second order lpf circuit. the opamp Tl082 has a gain bandwidth product of 4Mhz according to the datasheet. this circuit has a bandwidth of 414.25khz. can we determine the opamp' individual gain-bandwidth product from this plot ??
if yes , how do we find the answer ? bode ac.JPGopamp.JPG
 

Yes, it is possible. However, you have to accept some errors caused by a simplification.
That means: You have to assume a simplified opamp open-loop gain expression A(s)=GBW/s=wt/s (wt=transit frequency).
This simplification would lead to a second-order lowpass response. But as you can see - your simulation reveals that the phase response exceeds -180 deg.
This is, of course, caused by the higher-order opamp model. However, it can be expected that the error will be relatively small.
The procedure is as follows:
* Calculate the second-order lowpass function of the circuit using A(s)=wt/s and identify the expression for the pole frequency wp (which contains wt=GBW)
* From the simulation (Bode diagram) find the pole frequency wp. Assuming a second-order response, this frequency wp can be found for phase=-90 deg.
* Compare both wp values (calculated vs. simulated) and solve for wt=GBW.

For determination of the transfer function I have used a symbolic simulation program.
After some manipulations I arrived at

H(s)=Vo/Vin=1/D(s) and D(s)=1+s*(K/wt)+[s*(K/wt)]^2 with K=1+R2/R1

Comparing the denominator D(s) with the classical second-order form D(s)=1+s/(Qp*wp)+[s/wp]^2
we immediately find: wp=wt/K or wt=GBW=wp*K.

From your simulation:
Pole frequency wp is approximately 400...420 kHz (find exact value at phase=-90 deg) .
Thus: wt=GBW=wp*K=wp*(1+R2/R1).

Hope it helps.
 
Yes, it is possible. However, you have to accept some errors caused by a simplification.
That means: You have to assume a simplified opamp open-loop gain expression A(s)=GBW/s=wt/s (wt=transit frequency).
This simplification would lead to a second-order lowpass response. But as you can see - your simulation reveals that the phase response exceeds -180 deg.
This is, of course, caused by the higher-order opamp model. However, it can be expected that the error will be relatively small.
The procedure is as follows:
* Calculate the second-order lowpass function of the circuit using A(s)=wt/s and identify the expression for the pole frequency wp (which contains wt=GBW)
* From the simulation (Bode diagram) find the pole frequency wp. Assuming a second-order response, this frequency wp can be found for phase=-90 deg.
* Compare both wp values (calculated vs. simulated) and solve for wt=GBW.

For determination of the transfer function I have used a symbolic simulation program.
After some manipulations I arrived at

H(s)=Vo/Vin=1/D(s) and D(s)=1+s*(K/wt)+[s*(K/wt)]^2 with K=1+R2/R1

Comparing the denominator D(s) with the classical second-order form D(s)=1+s/(Qp*wp)+[s/wp]^2
we immediately find: wp=wt/K or wt=GBW=wp*K.

From your simulation:
Pole frequency wp is approximately 400...420 kHz (find exact value at phase=-90 deg) .
Thus: wt=GBW=wp*K=wp*(1+R2/R1).

Hope it helps.

Capture.JPG

I Substituted A=GB/s not GB/2 sorry for the typo in the pic

where alpha = r1/r1+r2

Sorry Mr.LvW, i dont clearly understand what u said. i derived the transfer function as shown in the picture. but how to find wp(cut off frequency) from the second equation. i can substitute 1/sqr(2) for v0/vi. but i dont know GBW and wp in the equation. please help me. thank u.
 
Last edited:

View attachment 80867
I Substituted A=GB/s not GB/2 sorry for the typo in the pic
where alpha = r1/r1+r2
Sorry Mr.LvW, i dont clearly understand what u said. i derived the transfer function as shown in the picture. but how to find wp(cut off frequency) from the second equation. i can substitute 1/sqr(2) for v0/vi. but i dont know GBW and wp in the equation. please help me. thank u.

Your result is identical to the transfer function I have found if you replace:
alpha=1/K and GB=wt.

Please note that you shouldn't write "G^2*B^2". GB is just an abbreviation for the "gain-bandwidth-product" which is (for unity gain compensated opamps like TL082) identical to the transit frequency wt.

Now to your question:
You have two transfer functions: The first one is calculated in terms of GB=wt (as shown above) and the second one is the general form - expressed by the classical pole data Qp and wp.
Please note, that the pole frequency is NOT identical to the cut-off frequency.
Now, you have only to compare both formulas if both are written in the same form (D(s)=1+a+s+b*s^2).
In detail: you compare the factors of s^2 which give you the formula for (1/wp)^2.
This leads to wp=wt*alpha=wt/K.
That's all you need. As I told you - the value of pole frequency wp for your circuit can be found at the frequency where the phase crosses -90 deg.
Now you can calculate the value for wt=GB.
Please note: You do NOT need the cut-off frequency. It does not help because it does not appear in the general transfer function form.

- - - Updated - - -

Remark: Don't forget that your BODE diagram is scaled in 1/sec=Hz - in contrast to the formulas above which contain w in rad/sec (w=2*Pi*f).
 
Your result is identical to the transfer function I have found if you replace:
alpha=1/K and GB=wt.

Please note that you shouldn't write "G^2*B^2". GB is just an abbreviation for the "gain-bandwidth-product" which is (for unity gain compensated opamps like TL082) identical to the transit frequency wt.

Now to your question:
You have two transfer functions: The first one is calculated in terms of GB=wt (as shown above) and the second one is the general form - expressed by the classical pole data Qp and wp.
Please note, that the pole frequency is NOT identical to the cut-off frequency.
Now, you have only to compare both formulas if both are written in the same form (D(s)=1+a+s+b*s^2).
In detail: you compare the factors of s^2 which give you the formula for (1/wp)^2.
This leads to wp=wt*alpha=wt/K.
That's all you need. As I told you - the value of pole frequency wp for your circuit can be found at the frequency where the phase crosses -90 deg.
Now you can calculate the value for wt=GB.
Please note: You do NOT need the cut-off frequency. It does not help because it does not appear in the general transfer function form.

- - - Updated - - -

Remark: Don't forget that your BODE diagram is scaled in 1/sec=Hz - in contrast to the formulas above which contain w in rad/sec (w=2*Pi*f).

thank u Mr.LvW. but theres a problem in my calc. as u said i found wp from the simulation plot by finding fp=302.26k hz(diagram) and finding wp=2*pi*fp.

the K value is 11 (given by 11k/1k). using these values i land up with gb or wt as
11*wp=20.8*10^6, which means the opamp has unity gain till 20.8Mhz. This is xtremely non-intuitive. kindly help .thank u.
 

thank u Mr.LvW. but theres a problem in my calc. as u said i found wp from the simulation plot by finding fp=302.26k hz(diagram) and finding wp=2*pi*fp.
the K value is 11 (given by 11k/1k). using these values i land up with gb or wt as
11*wp=20.8*10^6, which means the opamp has unity gain till 20.8Mhz. This is xtremely non-intuitive. kindly help .thank u.

Everything OK - except your interpretation:

The product (11*wp) gives wt=6.28*ft in rad/sec.
This leads to ft=20.8/6.28=3.3 MHz. And that sounds reasonable, does it not?
 
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